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Unformatted text preview: Version 028/AABDA – midterm 02 – Turner – (59130) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points After a 5 . 65 Ω resistor is connected across a battery with a 0 . 08 Ω internal resistance, the electric potential between the physical battery terminals is 7 V. What is the rated emf of the battery? 1. 13.5033 2. 10.4897 3. 10.1571 4. 7.09912 5. 12.9461 6. 13.175 7. 9.43299 8. 11.5814 9. 15.4682 10. 7.45337 Correct answer: 7 . 09912 V. Explanation: Let : R = 5 . 65 Ω , r = 0 . 08 Ω , and V = 7 V . The current drawn by the external resistor is given by I = V R = 7 V 5 . 65 Ω = 1 . 23894 A . The output voltage is reduced by the inter nal resistance of the battery by V = E − I r , so the electromotive force is E = V + I r = 7 V + (1 . 23894 A) (0 . 08 Ω) = 7 . 09912 V . 002 10.0 points A capacitor network with airfilled capacitors as shown below. 80 . 7 V 25 . 8 μ F 25 . 8 μ F 25 . 8 μ F 25 . 8 μ F b a c d When the top righthand capacitor is filled with a material of dielectric constant κ , the charge on this capacitor is increases by a fac tor of 1 . 24. Find the dielectric constant κ of the mate rial inserted into the top righthand capacitor. 1. 1.5641 2. 1.5 3. 1.98507 4. 2.1746 5. 2.50877 6. 2.27869 7. 2.0303 8. 1.53165 9. 3.0 10. 1.63158 Correct answer: 1 . 63158. Explanation: Let : C 1 = C = 25 . 8 μ F , C 2 = C = 25 . 8 μ F , C 3 = C = 25 . 8 μ F , C 4 = C = 25 . 8 μ F , E B = 80 . 7 V , and Q ′ = 1 . 5 Q . Version 028/AABDA – midterm 02 – Turner – (59130) 2 E B C 1 C 3 C 2 C 4 b a c d The capacitors C 3 and C 4 have nothing to do with this problems. In addition, the capac itances are all equal and their specific values are immaterial. Furthermore, the electric po tential of the battery is not required. C 1 = C 2 = C 3 = C 4 , where Q and Q ′ are the initial and final charges on C 2 and Q ′ Q ≡ α =ratio of final to initial charge on C 2 . We know the charges on C 1 and C 2 are the same. Initially, V ab = V 1 + V 2 = Q C 1 + Q C 2 = Q C + Q C = 2 Q C . (1) Therefore Q = 1 2 V ab C . After the dielectric material is inserted in C 2 , the capacitance becomes C ′ 2 = κC . There fore, V ab = V ′ 1 + V ′ 2 = Q ′ C 1 + Q ′ C ′ 2 = Q ′ C + Q ′ κC = κ + 1 κ Q ′ C , and using Eq. (1) and solving for Q ′ , we have 2 Q C = κ + 1 κ Q ′ C Q ′ = κ κ + 1 V ab C = κ κ + 1 2 Q Q ′ Q ≡ α = 2 κ κ + 1 = 1 . 24 . Solving for κ , we have κ = α 2 − α = 1 . 24 2 − 1 . 24 = 1 . 63158 . keywords: 003 10.0 points An airfilled cylindrical capacitor has a capac itance of 16 pF and is 5 . 9 cm in length....
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 Spring '08
 Turner
 Capacitance, Electric charge, Version 028/AABDA

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